$A$ tuning fork gives $5 \text{ beats per second}$ with a $33 \text{ cm}$ length of sonometer wire. If the length of the wire is shortened by $1 \text{ cm}$, the number of beats is still the same. The frequency of the fork is (in $\text{ Hz}$)

If the tension of a sonometer wire increases four times,then the fundamental frequency of the wire will increase by how many times?

Two uniform wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the $2^{\text{nd}}$ overtone of the $2^{\text{nd}}$ wire and the radius of the first wire is twice the radius of the $2^{\text{nd}}$ wire,then the ratio of the length of the first wire to the $2^{\text{nd}}$ wire is:

When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 \,cm$ or $100 \,cm$. Then the frequency of the tuning fork is (Consider that the tension in the wire is kept constant). (in $\,Hz$)

When tension $T$ is applied to a sonometer wire of length $l$, it vibrates with the fundamental frequency $n$. Keeping the setup same, when the tension is increased by $8 \,N$, the fundamental frequency becomes three times the earlier. The initial tension applied to the wire was: (in $\,N$)

$A$ body is suspended from a string of length $1 \ m$ and mass $2 \ g$. The mass of the body required to produce a fundamental mode of $100 \ Hz$ frequency in the string is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)